Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $r = \dfrac{-5q^2 - 55q - 140}{3q^2 + 30q + 72} \times \dfrac{q + 1}{q + 7} $
First factor out any common factors. $r = \dfrac{-5(q^2 + 11q + 28)}{3(q^2 + 10q + 24)} \times \dfrac{q + 1}{q + 7} $ Then factor the quadratic expressions. $r = \dfrac {-5(q + 4)(q + 7)} {3(q + 4)(q + 6)} \times \dfrac {q + 1} {q + 7} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { -5(q + 4)(q + 7) \times (q + 1)} { 3(q + 4)(q + 6) \times (q + 7)} $ $r = \dfrac {-5(q + 4)(q + 7)(q + 1)} {3(q + 4)(q + 6)(q + 7)} $ Notice that $(q + 4)$ and $(q + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-5\cancel{(q + 4)}(q + 7)(q + 1)} {3\cancel{(q + 4)}(q + 6)(q + 7)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $r = \dfrac {-5\cancel{(q + 4)}\cancel{(q + 7)}(q + 1)} {3\cancel{(q + 4)}(q + 6)\cancel{(q + 7)}} $ We are dividing by $q + 7$ , so $q + 7 \neq 0$ Therefore, $q \neq -7$ $r = \dfrac {-5(q + 1)} {3(q + 6)} $ $ r = \dfrac{-5(q + 1)}{3(q + 6)}; q \neq -4; q \neq -7 $